Ken: Echo out the query. Check it. Run it from the MySQL client.
What does the error message say? Does the table "109fh6" actually exist in the database? David # Ken responded: # # After adding echo mysql_error(); I get the same result. I tried changing # the query to include 109fh7 (a table which doesn't exist) and got the same # result as with 109fh6. Changing to 109fh5 does pull up that table. The # line to which the error message refers is while ($row = mysql_fetch_assoc # ($data_set)) That is what always come up when there is an error in the # query. ************** On Mon, 12 Dec 2005 14:13:10 -0500, Micah Stevens <[EMAIL PROTECTED]> wrote: > > You're getting an error, after the query, put: > > echo mysql_error(); > > to find out what's happening. > > On Monday 12 December 2005 11:05 am, [EMAIL PROTECTED] wrote: >> I made tiny changes to my php file and sql table and the table won't >> come >> up. I updated the table name (and php file name) from 109fh5 to 109fh6. >> In the table, I changed 6 cells, leaving a couple blank. Then I changed >> only the digit "5" to make it a "6" (109fh6) in the following: >> >> $get_data_query = "select rep, party, state, cd, minority, afr_am, >> asian, >> am_indian, hispanic, med_hsehold_income, poverty from 109fh6 order by >> $sort_field $sort_order"; >> >> Now I get "Warning: mysql_fetch_assoc(): supplied argument is not a >> valid >> MySQL result resource in" etc. >> >> I've done this many times without a problem (this is the 6th time in >> this >> sequence). What could be wrong after such a minor change? >> >> Ken > -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php